Subject:

You are correct about reflexivity and symmetry.

The main diagonal is all $1$’s, so the relation is reflexive, and the matrix is symmetric about the main diagonal, so the relation is symmetric.

A relation $R$ is antisymmetric if $$(a R b\text{ AND }bRa)\implies (a=b)\;.$$ For your first relation you have $1R3$ and $3R1$, and yet $3\ne 1$, so this relation is *not* antisymmetric.

For the same reason the second and fourth relations are not antisymmetric.

The third relation is not antisymmetric because $1R5$ and $5R1$, even though $1 \ne 5$. The fifth is not antisymmetric because $2R5$ and $5R2$, even though $2 \ne 5$.

Thus, *none* of the relations is antisymmetric. The visual test is to see whether there is a pair of $1$’s that are symmetric with respect to the main diagonal; if there are $-$ and we found such a pair of every one of these five relations $-$ then the relation is *not* antisymmetric.

There is no simple test for transitivity.

For the first relation, note that $2R3$ and $3R1$, but it’s not true that $2R1$, so the first relation is not transitive.

In the third relation $1R4$ and $4R2$, but it’s not true that $1R2$, so the relation is not transitive. In the fourth relation $2R3$ and $3R1$, but it isn’t true that $2R1$, so the relation is not transitive.

In the last relation $3R2$ and $2R1$, but it’s not the case that $3R1$, so the relation is not transitive.

The second relation is, as you say, transitive. One way to demonstrate this is to notice that in this relation every odd number in $\{1,2,3,4,5\}$ is related to every odd number, and every even number to every even number, but no odd number is related to any even number or vice versa.

## Binary Relations IIT JEE Main Mathematics Online Mock Test Problem 61(April 15 2018 Morning)

Thus, if $aRb$ and $bRc$, then either $a,b$, and $c$ are all odd, or they’re all even, and in either case it’s true that $aRc$.

answered Dec 30 '11 at 11:21

ScottBrian M. Scott

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